The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the :
(a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).
Explanation
Let (A = begin{pmatrix} a & b \ c & d end{pmatrix}).
(T : begin{pmatrix} a & b \ c & d end{pmatrix} = begin{pmatrix} a & b \ c & d end{pmatrix} begin{pmatrix} 3 & 5 \ 4 & 6 end{pmatrix})
(2a + 3b = 3 … (1))
(4a + 5b = 5 … (2))
(2c + 3d = 4 … (3))
(4c + 5d = 6 …. (4))
Solving (1) & (2) :
((1) times 2 : 4a + 6b = 6 … (5))
((5) – (2) : b = 1)
(2a + 3(1) = 3 implies 2a = 0; a = 0)
((3) times 2 : 4c + 6d = 8 … (6))
((6) – (5) : d = 2)
(2c + 3(2) = 4 implies 2c = -2; c = -1)
(a) (therefore A = begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix})
(b) Let the inverse of A be A(^{-1}).
(AA^{-1} = 1)
(begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} begin{pmatrix} w & x \ y & z end{pmatrix} = begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix})
(begin{pmatrix} y & z \ -w + 2y & -x + 2z end{pmatrix} = begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix})
(implies y = 1; z = 0)
(-w + 2y = 0 implies -w + 2 =0)
(-w = -2 implies w = 2)
(-x + 2z = 1 implies -x + 2(0) = 1)
(-x = 1 implies x = -1)
(therefore A^{-1} = begin{pmatrix} 2 & -1 \ 1 & 0 end{pmatrix})
(c) Let the point be (x, y).
(A : x to begin{pmatrix} 0 & 1 \ -1 & 2 end{pmatrix} begin{pmatrix} x \ y end{pmatrix} = begin{pmatrix} -1 \ 1 end{pmatrix})
(0 + y = -1 implies y = -1)
(-x + 2y = 1 implies -x + 2(-1) = 1)
(-x – 2 = 1 implies -x = 3)
(x = -3).
The point is (-3, -1).